# Explaining difference between improper integrals that converge and diverge

How would you explain the difference in the results given by integration of the two functions $$y = \frac{1}{1+x}$$ and $$y = \frac{1}{1+x^2}$$ ?

The graphs of these two function look so similar on positive x-values as x tends to infinity. In both cases y approaches to 0 and y = 0 is a horizontal asymptote in both graphs, but when you integrate with respect to x from 0 to infinity, the first graph has infinite area between the graph and the x axis, while the second graph has finite value π/2 as the area.

How would you explain this difference to students?

• You are likely to get the answer "ask this on math.stackexchange" unless you further edit the question to make it clearly a math education question, instead of a math question. Why isn't the answer to this question just "compute the two integrals?" Jun 22, 2022 at 18:11
• If you do this after infinite series, it may make more sense to the students. Jun 22, 2022 at 18:19
• One approaches $0$ faster than the other. Similar to: $\sum\frac{1}{n}$ diverges but $\sum\frac{1}{n^2}$ converges. Jun 22, 2022 at 18:51
• @SueVanHattum, I myself understood sums like 1/n versus 1/n^2 much more easily after learning the integral test! True, the notion of convergence of "indefinite" integrals is more complicated, but if we are optimists, it gives a clearer answer? :) Jun 22, 2022 at 19:31
• @JanakaRodrigo: Ah sorry, I always confuse the words "indefinite integral" and "improper integral" in English. Jun 23, 2022 at 12:03

Perhaps it's easier for students to see the reason for these results when you are looking at infinite series. Does the course you teach include this? (If not, you can still do it and set up a diagram showing that the infinite series is smaller than the diverging integral but still diverges, and also showing the converging series as bigger than its integral but still converging.)

It's easy to explain why you diverge when adding $$\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots$$. ($$\frac{1}{3}+\frac{1}{4}$$ is greater than $$\frac{1}{2}$$. So is the next group of 4 terms, and the next group of 8 terms, etc.)

For the other, you just need a convincing argument that it converges. Here's my best thought at the moment. The first 3 terms ($$\frac{1}{2}+\frac{1}{5}+\frac{1}{10}$$) add to less than 1. Ignore them, basically. The next 4 add to less than $$\frac{4}{16}=\frac{1}{4}$$, and the next 8 add to less than $$\frac{8}{64}=\frac{1}{8}$$, and so on. Which all adds to $$\frac{1}{2}$$. (I was surprised that this went to a geometric series when I did this. But I believe I have this right.)

• Nice example to demonstrate two types of behaviors when terms getting smaller and smaller in a series and this can be very much similar to what is happening closer to asymptotes in terms of area between the curve and the asymptote. Jun 23, 2022 at 15:52
• You're right about the geometric series, because$$\sum_{k=2^{n-1}}^{k=2^n-1}\tfrac{1}{1+k^2}<\sum_{k=2^{n-1}}^{k=2^n-1}\tfrac{1}{(2^{n-1})^2}=\tfrac{1}{2^{n-1}}.$$
– J.G.
Jun 24, 2022 at 14:04

I'd say: Don't trust your eyes, especially not when infinity is concerned and you are plotting in one small region. Sure, they look fairly similar when you are plotting over $$[1,10]$$. $$1/(1+x)$$ in blue, $$1/(1+x^2)$$ in red. Honestly, I think they look fairly different already, but even still.

Now try plotting the two over $$[10^{10}, 10^{11}]$$. Again, $$1/(1+x)$$ in blue, $$1/(1+x^2)$$ in red. The red one is pretty much gone compared to the blue.

After that, you pretty much have to go to proofs and establish bounds, etc.

• I think this is a dangerous approach to thinking about infinity. Who knows what happens on all the intervals you haven't glanced at? Jun 22, 2022 at 22:56
• @SueVanHattum depends what you are trying to show, right? You couldn't use these graphs to prove the area under the blue curve is infinity and the area under the red curve isn't. But if someone asks "how come they're not the same?" you could easily show them the blue curve vaguely seems to have more area under it Jun 23, 2022 at 8:56
• @SueVanHattum That's more or less the point I was trying to express with the last line. Otherwise, I endorse user253751 's comment.
Jun 23, 2022 at 12:50
• @Adam I would like to add a family of curves to this . Let's consider graphs of 1/x(xⁿ + 1) where n is positive number. When you integrate this from 1 to infinity, definite integral is ( ln2 )/n , value of this decreases as n increases but it is always finite value. All of the graphs of this family have X axis and Y axis as asymptotes but no stationary points when x is positive. If you find the area between Y axis and x = 1 ,that is infinite . Therefore when you consider the asymptote Y axis definite integral diverges but when you consider asymptote X axis it converge . Jun 24, 2022 at 1:05
• @JanakaRodrigo Sure, one could plot such a function as a first verification that $\frac{1}{x(x^n+1)}$ has a $1/x$--like singularity at $x=0$ and a $1/x^{n+1}$--like asymptote towards $x=\infty$. Then go back and do something more rigorous. I think, though, this shouldn't be something that you do every time. Just when you want to break false intuition or to do an "am I crazy?" check.
Jun 25, 2022 at 13:09

As Sue VanHattum suggests, it makes sense to shift gears to series first, for simplicity.

I would also do two examples of series with decreasing terms, one where it's obvious that it converges, and the other where it's obvious that it diverges. Good examples would be:

0.1 + 0.01 + 0.001 + ...

1.1 + 1.01 + 1.001 + ...

Next you could do an example whose terms approach zero, but clearly not quickly enough to converge. For example:

1 + 1/2 + 1/2 + 1/3 + 1/3 + 1/3 + ...

(where the $$1/n$$ term is repeated $$n$$ times).

Sue VanHattum's example of the harmonic series is fine, but the argument is tricky. I'd avoid trickiness when introducing difficult new ideas.

• My series are, in some sense, the same thing as the integral. But I like your examples. Jun 24, 2022 at 5:17

To me (a dumb troll), the way to intuitively think about it is 0 versus infinity. Infinity makes things huge. Zero makes them small. So, if you have an asymptotic approach you have a huge (to forever) amount...infinite like. But with very small parts. So. If the "infinity is winning", you don't converge. If the zero is winning, you do. And it's like a battle of rock and hard place. And sometimes one wins and sometimes the other does. If the "zero-ing" is happening faster than the "infinity-ing", you get something that converges. If the infinitying is happening faster, you get an integral that blows up.

• 0 times infinity, who wins? Perfect image, and lots of good examples to go with it. Jun 22, 2022 at 21:02
• That "race" between asymptotes and infinity is also likely to be a familiar concept if they've already covered limits and L'Hospital's Rule. Jun 23, 2022 at 11:42

My feeling is that if students understand the fundamental theorem of calculus, they should not have too much trouble seeing where such results come from. And if they don't understand the fundamental theorem of calculus, that's a far more central topic than improper integrals, and time would be better spent shoring up understanding the fundamental theorem than on improving understanding of improper integrals.

Assuming the fundamental theorem is well in hand, here are a few thoughts. Your integrals are similar to $$\int_1^\infty x^{-2}\,dx \qquad \text{and}\qquad\int_1^\infty x^{-1}\,dx,$$ the first of which equals $$1$$ and the second of which is infinite. Students should be able to see that this blow-up doesn't happen "all of a sudden" as the exponent is changed from $$-2$$ to $$-1$$, but rather, emerges in a sensible way as the exponent is continuously increased from $$-2$$ to $$-1$$. Let $$N$$ be a positive real number and evaluate the analogous integrals with exponent $$-\left(1+\frac{1}{N}\right)$$: $$\int_1^\infty x^{-\left(1+\frac{1}{N}\right)}\,dx=\lim_{A\to\infty}-N(A^{-\frac{1}{N}}-1^{-\frac{1}{N}})=N.$$ As $$N$$ grows larger and larger, the exponent gets closer to $$-1$$ from below, and the value of the integral grows larger and larger without bound.

It is interesting to look at what happens as $$N$$ is doubled. For $$N=1,2,4,8,16,\ldots$$ the exponent is $$-2$$, $$-\frac{3}{2}$$, $$-\frac{5}{4}$$, $$-\frac{9}{8}$$, $$-\frac{17}{16}$$. As the area is $$N$$, it is doubling as well. Here's how some of the curves look. I think I would be hard pressed to tell apart the top two curves if they were shown as separate graphs. That the area under the top curve is twice that of the curve second from the top is not something one can see from the graph, but is a testament to the dramatic effect of integrating all the way out to infinity. For similar reasons I don't think it is possible to visually distinguish curves with divergent integrals from curves with convergent ones.

In a comment you asked about the $$y$$-axis asymptote of the integrands $$\frac{1}{x(x^n+1)}$$. (You noted that for $$n$$ positive these integrals between $$0$$ and $$1$$ are divergent, although they are convergent between $$1$$ and $$\infty$$.) Let's look at something simpler. The graph of $$y=\frac{1}{x}$$ is symmetric about the line $$y=x$$, so the divergence of the integral from $$0$$ to $$1$$ is the same geometric fact as the divergence of the integral from $$1$$ to $$\infty$$. (Well the integral from $$0$$ to $$1$$ includes an extra $$1\times1$$ square, but is otherwise the same shape as the integral from $$1$$ to $$\infty$$, just reflected.)

More generally, if the graph of $$y=\frac{1}{x^n}$$ is reflected about $$y=x$$ we get the graph of $$y=\frac{1}{x^{1/n}}$$. Now $$\int_1^\infty \frac{1}{x^n}\,dx$$ is divergent for $$n\le1$$ and convergent for $$n>1$$. So $$\int_1^\infty \frac{1}{x^2}\,dx$$ is convergent while $$\int_1^\infty \frac{1}{x^{1/2}}\,dx$$ is divergent. On the other hand $$\int_0^1 \frac{1}{x^2}\,dx$$ represents the same area as $$\int_1^\infty \frac{1}{x^{1/2}}\,dx$$ apart from a $$1\times1$$ square, and hence is divergent. Likewise $$\int_0^1 \frac{1}{x^{1/2}}\,dx$$ represents the same area as $$\int_1^\infty \frac{1}{x^2}\,dx$$ apart from that $$1\times1$$ square again and hence is convergent.

Your integral $$\int_0^1\frac{1}{x(x^n+1)}\,dx$$ looks very much like $$\int_0^1\frac{1}{x}\,dx$$, at least for the part near $$x=0$$, which is where the blow-up occurs, so the divergence is not surprising.

• that specific example of family of graphs I mentioned because all of them have both convergent and divergent definite integrals relevant to area between the graph and asymptotes . 1/x has both integrals divergent and when you multiply it by 1/(xⁿ +1) values of 1/x will be decreased because 1/(xⁿ +1) is less than 1 for positive values of x. When x is very large decreasing becomes significant and that's why definite integral becomes convergent but when x is less than 1,decreasing does not create big difference since 1/x is very large, so definite integral remain as divergent. Jun 25, 2022 at 6:57
• Do you consider my examples with integrand $1/x^n$ and $n\ne1$ to be similar to your example in that the area between the curve and one asymptote is finite and the area between the curve and the other asymptote is infinite? They seem very similar to me. Jun 27, 2022 at 16:48
• yes they are similar except when you express 1/ x ⁿ , as a product of 1/x and 1/ x ⁿ⁻¹ . Because here both approach infinity as x tends to 0 . Jun 27, 2022 at 17:40
• I'm not sure what you're getting at here. Why is it important how the integrand can be written as a product? Jun 27, 2022 at 23:11
• it's not very important but I just wanted to compare in that way because my examples can be expressed as a product. Jun 27, 2022 at 23:35